Question

The value of a for which system of equations, \[{a^3}x + {(a + 1)^3}y + {(a + 2)^3}\;z = 0,\;\] \[ax + \left( {a + 1} \right)y + \left( {a + 2} \right)z = 0,\] \[x + y + z = 0\] has non zero solutions is :
A) -1
B) 0
C) 1
D) 2

Answer 1

Hint: The determinant of a 3 x 3 matrix A,
$A = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right)$
is defined as
$A = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right) = ({a_{11}}{a_{22}}{a_{33}} + {a_{12}}{a_{23}}{a_{31}} + {a_{13}}{a_{21}}{a_{32}}) - ({a_{31}}{a_{22}}{a_{13}} + {a_{32}}{a_{23}}{a_{11}} + {a_{33}}{a_{21}}{a_{12}})$
An easy method for calculating 3 X 3 determinants is found by rearranging and factoring the terms given above to get
\[A = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right) = {a_{11}}({a_{22}}{a_{33}} - {a_{32}}{a_{23}}) - {a_{21}}({a_{33}}{a_{12}} - {a_{13}}{a_{32}}) + {a_{31}}({a_{12}}{a_{23}} - {a_{22}}{a_{13}})\]

Complete step-by-step answer:
We'll start with a 3 x 3 matrix A, and try to find its determinant |A|.
The system of equation has a non-zero solution
Hence, \[\left( {\begin{array}{*{20}{c}}
  {{a^3}}&a&1 \\
  {{{(a + 1)}^3}}&{\left( {a + 1} \right)}&1 \\
  {{{(a + 2)}^3}}&{(a + 2)}&1
\end{array}} \right) = 0\]

Use row subtraction to make the matrix easier. If you take the values of one row and subtract them to a different row, the determinant of the matrix does not change. The same is true of columns. You can do this repeatedly — or multiply the values by a constant before adding — to get as many zeroes in the matrix as possible. This can save you a lot of time.
\[{R_1} \to {R_1} - {R_{3}},{R_{2}} \to {R_2} - {R_3}\]
$ \Rightarrow \left( {\begin{array}{*{20}{c}}
  {{a^3} - {{(a + 2)}^3}}&{ - 2}&0 \\
  {{{(a + 1)}^3} - {{(a + 2)}^3}}&{ - 1}&0 \\
  {{{(a + 2)}^3}}&{(a + 2)}&1
\end{array}} \right) = 0$
Expand along R3
\[ \Rightarrow + \left\{ {{a^3} - {{(a + 2)}^3}} \right\}\left( {\begin{array}{*{20}{c}}
  { - 1}&0 \\
  {(a + 2)}&1
\end{array}} \right) - ( - 2)\left( {\begin{array}{*{20}{c}}
  {{{(a + 1)}^3} - {{(a + 2)}^3}}&0 \\
  {{{(a + 2)}^3}}&1
\end{array}} \right) + 0\left( {\begin{array}{*{20}{c}}
  {{{(a + 1)}^3} - {{(a + 2)}^3}}&{ - 1} \\
  {{{(a + 2)}^3}}&{(a + 2)}
\end{array}} \right)\]
The process is called an expansion of the first row because as you can see in equation, all of the elements from the first row of the original 3x3 matrix remain as main factors in the expansion to be solved for. All of the 2x2 matrices in the expansion are what we call "secondary matrices", and they can be easily resolved using the equation on the determinant of a 2x2 matrix.
$ \Rightarrow \left\{ {{a^3} - {{(a + 2)}^3}} \right\}( - 1) + 2\left\{ {{{(a + 1)}^3} - {{(a + 2)}^3}} \right\} + 0$
\[
   \Rightarrow \left[ {\left\{ { - {a^3} + {{(a + 2)}^3}} \right\} + 2{{(a + 1)}^3} - 2{{(a + 2)}^3}} \right] = 0 \\
   \Rightarrow \left[ { - {a^3} - {{(a + 2)}^3} + 2{{(a + 1)}^3}} \right] = 0 \\
   \Rightarrow \left[ { - {a^3} - ({a^3} + 8 + 6{a^2} + 12a) + 2{a^3} + 2 + 6{a^2} + 6a} \right] = 0 \\
   \Rightarrow \left[ { - {a^3} - {a^3} - 8 - 6{a^2} - 12a + 2{a^3} + 2 + 6{a^2} + 6a} \right] = 0 \\
   \Rightarrow \left[ { - 6 - 6a} \right] = 0 \\
   \Rightarrow a = - 1 \\
 \]
So, option (A) is the correct answer.

Note: Be very careful when substituting the values into the right places in the formula and to keep track of all negative signs when evaluating determinants. Common errors occur when students become careless during the initial step of substitution of values. Work carefully, writing down each step as in the examples. Skipping steps frequently leads to errors in these computations.
In addition, take your time to make sure your arithmetic is also correct. Otherwise, a single error somewhere in the calculation will yield a wrong answer in the end.
Exactly the same answer would be found using any row or column of the matrix.